带有await的firebase on('value')不能按预期工作

Question

我正在寻找构建函数,等待所有值都打开('value'),然后转到下一行,换句话说是异步函数.

 let upcomingGamesList = await firebase.database().ref('UpcomingGames').on('value', snapshot => {
        upcomingGamesList = snapshot.val()
        console.log('upcoming t1',upcomingGamesList)
        return upcomingGamesList
    })
    console.log('upcoming t2',upcomingGamesList)
    let upcomingPreferences = upcomingGamesList.map(async(game) => {
        console.log(game)
        let GameId = game.GameId
        await firebase.database().ref(`GameNotificationPreferances/${GameId}`).orderByKey().equalTo(UserStore.user.uid).once('value', snapshot => {
            if (snapshot.val() != null || snapshot.val() != undefined) {
                conosle.log(snapshot.val())
            } else {
                console.log('not value')
            }
        })
        console.log(game)
    })

发生了什么是即将到来的t2

        console.log('upcoming t2',upcomingGamesList)

在即将到来的t1之前打印

    console.log('upcoming t2',upcomingGamesList)

但是我在这个函数中使用了await

  let upcomingGamesList = await firebase.database().ref('UpcomingGames').on('value', snapshot => {
        upcomingGamesList = snapshot.val()
        console.log('upcoming t2',upcomingGamesList)
        return upcomingGamesList
    })

它应该等到它完成然后转到下一行

我想等到函数完成后再使用我所做的更改获取updatedList

 let upcomingGamesList = await firebase.database().ref('UpcomingGames').on('value', async(snapshot) => {
        upcomingGamesList = snapshot.val()
        updatededList = await upcomingGamesList.map(async(game) => {
            let GameId = game.GameId
            await firebase.database().ref(`GameNotificationPreferances/${GameId}`).orderByKey().equalTo(UserStore.user.uid).once('value', async(snapshot) => {
                if (snapshot.val() != null || snapshot.val() != undefined) {
                    game['reminderPressed'] = true;
                } else {
                    game['reminderPressed'] = false
                }
                console.log('GameId:',GameId, 'GameDetails:',game)
                return ({...game})

            })

        })


    })
    console.log('the updatedList is',updatededList)

Answer 1

Firebase的on()功能将持续监听您呼叫它们的位置.这意味着他们可以多次为您提供数据.由于一个Promise只能解决一次,on()不会返回一个承诺.因此它不能与async/ 一起使用await.

在这种情况下,它看起来像你想要使用once(),它几乎相同,但只会传递一次结果(因此返回一个承诺):

let upcomingGamesList = await firebase.database().ref('UpcomingGames').once('value', snapshot => {
    upcomingGamesList = snapshot.val()
    console.log('upcoming t2',upcomingGamesList)
    return upcomingGamesList
})